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$X \sim N(0,2) \quad Y \sim N(0,1) \quad Cov(X,Y) = 0.5 \quad Z = X + XY + Y$

find:

1) $E(Y|X)$ and $Var(Y|X)$

2) $E(Z|X)$ and $Var(Z|X)$

3) $E(Z)$ and $Var(Z)$

4) $Cov(Z,Y|X)$

5) joint distribution of $Z$ and $Y$ conditional on $X$

I'm stuck from question 3) onwards since I'm not able to compute $Var(Z)$.

Is there a textbook, possibly free and online, explaining these issues and providing exercises WITH SOLUTIONS?

1)

$E(Y|X) = E(Y) + \frac{Cov(X,Y)}{Var(X)}(X - E(X)) = 0.25 X$

$Var(Y|X) = Var(Y) - \frac{Cov(X,Y)^2}{Var(X)} = 1 - \frac{0.5^2}{2} = 0.875$

2)

$E(Z) = E(X) + E(XY) + E(Y) = 0 + Cov(X,Y) + E(X)E(Y) + 0 = 0.5$

$Var(Z) = Var(X) + Var(XY) + Var(Y) + 2 Cov(X,XY) + 2 Cov(XY,Y) + 2 Cov(X,Y)$

I am not sure this is the right way of computing $Var(Z)$ and I'm not able to compute $Cov(X,XY)$

FrancescoVe
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    Unless I missed something I'm not sure there's enough information to answer 5. It would seem to require an assumption not given. – Glen_b Oct 31 '17 at 13:55
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    If you are stuck on question 3, why do you feel it necessary to ask 5 questions – wolfies Oct 31 '17 at 14:11
  • If you have answers already for 1) and 2), they most likely are based on an _unstated_ assumption about the _joint distribution_ of $X$ and $Y$. Could you _edit_ your question to make this assumption explicit? If you have no clue as to what might the assumption, see the answers to [this question](https://stats.stackexchange.com/q/30159/6633). – Dilip Sarwate Oct 31 '17 at 15:14
  • @wolfies I edited my question (added "onwards") – FrancescoVe Oct 31 '17 at 15:59
  • Unfortunately I have no additional information. This is an exercise from an econometrics exam that I am unable to solve. – FrancescoVe Oct 31 '17 at 16:01
  • It likely assumes $(X,Y)$ has a bivariate Normal distribution. Without that assumption the expectations of $(XY)^2$, $XY^2$, and $X^2Y$ cannot be computed--and they are a key part of the variance of $Z$. Perhaps this observation will now let you proceed with the rest of the problem? If not, please explain *why* you cannot compute the variance of $Z$. What formulas or definitions have you tried to apply and what obstacle(s) did you encounter? – whuber Oct 31 '17 at 16:38
  • When you're unsure of a formula, go back to definitions. In this case you might consider $$\operatorname{Var}(Z)=E(Z^2)-E(Z)^2=E(X^2+2X^2Y+2XY+2XY^2+2X^2Y^2)-(E(X)+E(XY)+E(Y))^2$$ and go on from there. – whuber Oct 31 '17 at 19:15

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